Re: How to pass a string as url's parameter

You probably don't want to use the name directly, you should probably look at creating a slug from the name and store it in a SlugField on your model.

https://docs.djangoproject.com/en/1.8/ref/models/fields/#slugfield
https://docs.djangoproject.com/en/1.8/ref/utils/#module-django.utils.text

Your URL regex already looks correct to capture a slug.

At that point it's just a matter of pulling the object via the slug in your view:

fruit = get_object_or_404(Fruit, slug=fruit_name)

Of course that's assuming that your SlugField is named 'slug'. Make sure to read up on slugs since they need to be unique, and generating them may be a challenge.

-James

On Aug 13, 2015 12:18 PM, "I. Dié" <idie.levano@gmail.com> wrote:

Hi evrybody,

Here is my model:

# fruits/models
    Class Fruit (models.Model):
            name =  models.CharField(max_length = 33,
                               choices = NAME_CHOICES, default = MANGO)
           # and so on...

# fruits/views
def fruit(request, fruit_name):
    fruit = get_object_or_404(Fruit)
    return render(request, 'fruits/fruit.html', {'fruit': fruit})

# fruits/urls
urlpatterns = [
    url(r'^$', views.index, name="index"),
    url(r'^(?P<fruit_name>[-\w]+)/$', views.fruit, name='fruit'),
]

When I used fruit_id, everything was ok.
But I want to know how to use Fruit.name (a string) as url's parameter instead of fruit_id like this: myverybigmarket/fruits/mango better than myverybigmarket/fruits/1

Ps: this is my first python/django project.

Thanks

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